Question

Sol. [Key Concept: Three normals from $(c,0)$ to $y^{2} = x$: one is the axis; the other two slopes satisfy a quadratic, and perpendicularity ($m_{1}m_{2} = - 1$) fixes $c$.]

Equation of normal at $\left( \frac{1}{4}t^{2},\frac{1}{2}t \right)$ is

$$y = - tx + \frac{1}{2}t + \frac{1}{4}t^{3} $$Put ( $c,0$ )

$$\left( \frac{1}{4}t_{2}^{2},\frac{1}{2}t_{2} \right)$$

$${\Rightarrow \frac{1}{4}t^{3} + \left( \frac{1}{2} - c \right)t = 0 }{\Rightarrow t^{2} - (4c - 2) = 0 \smallsetminus_{t_{2}}^{t_{1}} \Rightarrow 4c - 2 > 0 \Rightarrow c > \frac{1}{2}}$$ and $t_{1}t_{2} = - 1$

$$\Rightarrow \ 4c - 2 = 1 \Rightarrow c = 3/4 $$